Yes this one is well known. Often for finding a bijection it suffices to compare the reasons why the two interpretations satisfy the fundamental recurrence $C_{n+1}=\sum_{i=0}^nC_i\,C_{n-i}$ for Catalan numbers. For $n+1$-leaf binary trees, which is what combining $n+1$ atoms using $n$ applications of a binary operator amounts to, this is quite clear: for $C_{n+1}$ one is looking at trees with $n+1$ internal nodes one of which is the root; if it has a left subtree with $i$ internal nodes its right subtree has $n-i$ internal nodes, and the value of $i$ together with the choice of such subtrees determines the tree. For $2n$-step Dyck paths it is slightly less natural since an asymmetric choice is necessary to decompose a path in sub-paths. Still there is a fairly obvious way to do this: the first step of a $2(n+1)$ step path is always an up-step (opening parenthesis), and there is a unique matching down-step (closing parenthesis) where we first re-descend to level$~0$. Then there are $2i$ steps in between them that form a Dyck path, and $2(n-i)$ steps left after the matching down-step, for some $0\leq i\leq n$. Repeating this decomposition constructs a binary tree associated to a Dyck path. For instance with $n=3$ and the path corresponding to $(())()$ one finds $i=1$ (the first descent to level$~0$ is at step$~4$) and the unique binary tree with $4$ leaves and the root in the middle: $(a*b)*(c*d)$. Note that the left-right symmetry is not preserved. Here are all $5$ cases with $n=3$:
$$
\matrix{(~)~(~)~(~) & (~)~(~(~)~)~&(~(~)~)~(~) & (~(~)~(~)~)&(~(~(~)~)~)\\
a*(b*(c*d))&a*((b*c)*d) &(a*b)*(c*d) & (a*(b*c))*d & ((a*b)*c)*d}
$$
Here is a maybe somewhat easier to see way to state this correspondence. First write the binary tree expression using symbols '[', '$*$', ']' together with atoms, so that each '$*$' has a pair '[' and ']' that surrounds its own operands, so for instance $a*((b*c)*d)$ becomes $[~a*[~[~b*c~]*d~]~]$ (note the outer brackets). Then substitute '(' for '[', and ')' for '$*$', and drop both the atoms and the symbols ']', to get a balanced list of parentheses, in the example $(.)((.)..)...$ where the dots mark positions where symbols were dropped.
I find it a bit easier to think in terms of paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that consist of $n$ up-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell+1\rangle$) and $n$ down-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell-1\rangle$). An up-step in this version corresponds to a step to the right in your version, and a down-step corresponds to a step upwards in your version. Your boundary condition becomes a requirement that my path not drop below the line $y=-z$.
We can use a slight modification of one of the usual arguments for counting the paths that donโt drop below the line $y=0$.
As in your version, there are altogether $\binom{2n}n$ paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$, and the problem is to count the โbadโ ones, i.e., the ones that do drop below the line $y=-z$. Suppose that we have a bad path $\pi$. There is a first point at which $\pi$ reaches the line $y=-z-1$; if it has made $u$ up-steps at that point, it must have made $u+z+1$ down-steps and so have reached the point $\langle 2u+z+1,-z-1\rangle$. Reflect the remainder of $\pi$ (i.e., the part to the right of this point) in the line $y=-z-1$. That part of $\pi$ has $n-u$ up-steps and $n-u-z-1$ down-steps, so its reflection has $n-u$ down-steps and $n-u-z-1$ up-steps. This means that it must end at the point
$$\langle 2u+z+1,-z-1\rangle+\langle2n-2u-z-1,-z-1\rangle=\langle 2n,-2z-2\rangle\;.$$
Conversely, any path from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$ must hit the line $y=-z-1$, and if we reflect the part of it to the right of that intersection in the line $y=-z-1$, we get a path from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that drops below the line $y=-z$. Thus, we have a bijection between our bad paths and all paths from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$. Each of these paths has $n-z-1$ up-steps and $n+z+1$ down-steps, so there are $\binom{2n}{n+z+1}$ of them. Thus, there are
$$\binom{2n}n-\binom{2n}{n+z+1}=\binom{2n}n-\binom{2n}{n-z-1}$$
good paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$.
Best Answer
A picture will help to visualize the bijection described in this post : please take your pencil.
First, the setting : it will be easier to view Dyck paths as (ordered rooted) trees, using the traditional device of the contour function (I put a ref below). A peak is then a leaf of the tree (a vertex without children). Also, Dyck paths being of length 2๐, the trees will have n edges, so that the minimal number of leaves will be 1 and the maximal number will be ๐. Call brothers of a vertex the children of the same father (by convention, we discard the vertex itself).
Your claim is that, among trees on $n$ edges, there is as many with $k$ leaves as with $n+1-k$ leaves.
The bijection goes as follows : to a (rooted) tree on $n$ edges, we will associate another (rooted) tree with $n$ edges, by keeping the same vertex set (also the same root) and changing the edge set. To that end, it is enough to define the father of every vertex distinct from the root in the new tree, let's call it the new father.
Consider a vertex distinct from the root and look for the most recent ancestor of that vertex that has a brother to its right : this will the vertex itself if it has a brother to its right, or the father of this vertex if the vertex itself has no brother to its right but its father has, and so on...
In case there is such an ancestor : the new father is the brother immediately to the right of that ancestor.
In case there is no such ancestor : the new father is the root of the tree.
The converse map is obtained by replacing the word "right" by the word "left" in the above description.
I'll let you check this works with respect to the property you have in mind for the leaves : a possible proof is by recurrence on the number of leaves (it is easy to see it works for one leave, then one looks at the shape of trees with two leaves, and from there, the idea os the recurrence should be quite clear)
(This is closely related, but in a subtle way distinct from what is called rotation bijection in computer science, see Flajolet and Sedgewick on p.73.)
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The reference on the contour function (in French sorry):
https://fr.wikipedia.org/wiki/Arbre_de_Galton-Watson#Processus_de_contour.
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From far, the question of MathSE on which I spent the most time. Perhaps it is known in combinatorics, I have done no research on the topic.